Difference between revisions of "GPU610/NullPointerException"
(→Assignment 1) |
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while (count < n){ | while (count < n){ | ||
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flag = 1; | flag = 1; | ||
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i = 1; | i = 1; | ||
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checkKey = int((sqrt (prime))+1); | checkKey = int((sqrt (prime))+1); | ||
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//checks prime numbers for divisible values | //checks prime numbers for divisible values | ||
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while(primes[i] < checkKey and flag){ | while(primes[i] < checkKey and flag){ | ||
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if ((prime % primes[i])==0){ | if ((prime % primes[i])==0){ | ||
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flag = 0; | flag = 0; | ||
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} | } | ||
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i++; | i++; | ||
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} | } | ||
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if(flag){ | if(flag){ | ||
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primes[count] = prime; | primes[count] = prime; | ||
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count++; | count++; | ||
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} | } | ||
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prime +=2; | prime +=2; | ||
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} | } | ||
Revision as of 15:28, 7 February 2013
GPU610/DPS915 | Student List | Group and Project Index | Student Resources | Glossary
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Progress
Assignment 1
- Philip Aziz - Calculating Prime Numbers
Only the inner loop would be possible to Parellelize but would still require to validate 1 prime number at a time due to need to validate against previous prime numbers.
while (count < n){
flag = 1;
i = 1;
checkKey = int((sqrt (prime))+1);
//checks prime numbers for divisible values
while(primes[i] < checkKey and flag){
if ((prime % primes[i])==0){
flag = 0;
}
i++;
}
if(flag){
primes[count] = prime;
count++;
}
prime +=2;
}
| n | Elapsed Time |
|---|---|
| 10,000 | 0 |
| 100,000 | 1 |
| 1,000,000 | 8 |
| 10,000,000 | 220 |
| 100,000,000 | 6200 |
- Saad Mohammad -
- Natesh Mayuranathan -
